Every map $S^1 \rightarrow X$ extends to a map $D^2 \rightarrow X$ implies $\pi_1(X,x_0)$ is trivial

Question

Every map $S^1 \rightarrow X$ extends to a map $D^2 \rightarrow X$ implies $\pi_1(X,x_0)$ is trivial

I know this has been posted before, but I was hoping somebody could verify my proof for me:

Let $[f] \in \pi(X,x_0)$

Then we have $f$ is a map:

$f: I \rightarrow X$
such that
$f(0)=f(1)$

So we can consider $f$ a map from $S^1$ into $X$:

$f: S^1 \rightarrow X$

this map extends to $D^2$ by hypothesis:

$\tilde{f}: D^2 \rightarrow X$ where $S^1 = \delta D^2$

Define a homotopy $H: D^2 \times I \rightarrow X$ by

$((r,\theta),s) \rightarrow \tilde{f}(sr,\theta)$

Then we have:

$H((r,\theta),1)=\tilde{f}(r,\theta)$

$H((r,\theta),0)=\tilde{f}(0,\theta)=c_0$

And we can define $F(\theta,s)=H((1,\theta),s): S^1 \times I \rightarrow X$

So that:

$F(\theta,0)=H((1,\theta),0)=c_0$

$F(\theta,1)=H((1,\theta),1)=f(\theta)$

So that $F$ is a homotopy between $f$ and the constant map, $c_0$, and therefore $[f]=[c_0]$

Answer 1

Defining $F(z,t)=\tilde{f}(tz)$ is not enough because this maps doesn't fixe the base point. Instead you can define $F(z,t)=\tilde{f}((1-t)z+t)$ which satisfies $F(1,\cdot)=x_0$.

Edit: I just add a picture to visualize the homotopy.